@zepdrix does this one only simplify down to \[\int\limits_{}^{}\frac{ x^2-x+6 }{ x(x^2+3) }dx\] or am I making a silly mistake somewhere again. If that is as far as it factors down does the (x^2+3) change anything?

\[\large \frac{x^2-x+6}{x(x^2+3}\qquad = \qquad \frac{A}{x}+\color{royalblue}{\frac{Bx+C}{x^2+3}}\]
With partial fractions, you're top part will be one degree less than the bottom.
See how the first fraction is a constant over x^1?
Our second fraction will be `linear` on top, since the bottom is a quadratic.